Business Mathematics MTH-367

Lecture 15

Chapter 11

The Simplex and Computer Solutions Methods

continued

Last Lecture’s Summary

• Covered Sec. 11.1: Simplex Preliminaries• Requirements of Simplex Method• Requirements for different types of

constraints• Feasible Solution, Basic Solution, and

Basic Feasible Solution• Incorporating objective function

Today’s topics

• The Simplex Method for Maximization problem with all ≤ constraints

• Maximization problem with mixed constraints

Incorporating the Objective Function (Recall)

• Given the LP problem

Maximize z = 5x1 + 6x2

subject to 3x1 + 2x2 ≤ 120 (1)

4x1 + 6x2 ≤ 260 (2)

x1 , x2 ≥ 0

• The required form by simplex method is given as follows.

z – 5x1 – 6x2 – 0S1 – 0S2 = 0 (0)

3x1 + 2x2 + S1 = 120 (1)

4x1 + 6x2 + S2 = 260 (2)

• Since we are particularly concerned about the value of z and will want to know its value for any solution, z will always be a basic variable.

• The standard practice, however, is not to refer to z as a basic variable. The terms basic variable and non-basic variable are usually reserved for other variables in the problem.

Incorporating the Objective Function (recall)

• The simplex operations are performed in a tabular format.

• The initial table, or tableau, for our problem is shown in the table below.

• Note that there is one row for each equation and the table contains the coefficients of each variable in the equations and bi column contains right hand sides of the equation.

RULE: 1 OPTIMALITY CHECK IN MAXIMIZATION PROBLEM

In a maximization problem, the optimal solution has been found if all row (0) coefficients for the variables are greater than or equal to 0. If any row (0) coefficients are negative for non-basic variable, a better solution can be found by assigning a positive quantity to these variables.

RULE: 2 NEW BASIC VARIABLE IN MAXIMIZATION PROBLEM

In a maximization problem the non-basic variable which will replace a basic variable is the one having the most negative row (0) coefficient. “Ties” may be broken arbitrarily.

RULE: 3 DEPARTING BASIC VARIABLE

The basic variable to be replaced is found by determining the row i associated with

min ---- i = 1, . . . . m

Where aik > 0. In addition to identifying the departing basic variable, the minimum bi/aik value is the maximum number of units which can be introduced of the incoming basic variable.

bi

aik

Summary of simplex prodedure

We summaries the simplex procedure for maximization problems having all (≤) constraints.

First, add slack variables to each constraint and the objective function and place the variable coefficients and right-hand-side constants in a simplex tableau:

1- Identify the initial solution by declaring each of the slack variables as basic variables. All other variables are non-basic in the initial solution.

2- Determine whether the current solution is optimal by applying rule 1 [Are all row (0) coefficients ≥ 0?]. If it is not optimal, proceed to step 3.

Summary of Simplex procedure cont’d

3- Determine the non-basic variable which should become a basic variable in the next solution by applying rule 2 [most negative row (0) coefficient].

4- Determine the basic variable which should be replaced in the next solution by applying rule 3 (min bi/aik ratio where aik > 0)

5- Applying the Gaussian elimination operations to generate the new solution (or new tableau). Go to step 2.

Summary of Simplex procedure cont’d

Solve the following linear programming problem using the simplex method.

Maximize z = 5x1 + 6x2

subject to 3x1 + 2x2 ≤ 120

4x1 + 6x2 ≤ 260

x1 , x2 ≥ 0

Example

Rewriting the problem in standard form with slack variables added in, we have the following:

Maximize z = 5x1 + 6x2

subject to

x1 , x2 , S1 , S2 ≥ 0

• The objective function should be restated by moving all variables to the left side of the equation.

• The initial simplex tableau is shown in the following table.

Basic Variables

Key Column

Ratioz x1 x2 s1 s2 bi

1 –5 –6 0 0 0

S1 0 3 2 1 0 120

S2 0 4 6 0 1 260Step 1 in the initial solution x1 and x2 are non-basic

variable having values equal to 0. The basic variables are the slack variables with S1 = 120, S2 = 260, and z = 0.

Step 2 Now in table 1, we look for row(0) and observe that the current solution is optimal or not. Since all row (0) coefficients are not greater than or equal to 0, the initial solution is not optimal.

Step 3 The most negative coefficient in row (0) is –6, and it is associated with x2. So x2 column is the key column. Thus, x2 will become a basic variable in the next solution and its column of coefficients becomes the key column.

Step 4 In order to find the basic variable, which should be replaced in the next solution, we will look for the least ration in the last column.

In order to find the least ratio, we will divide the values of R.H.S by the corresponding values of the key column.

This ratio should be a real positive number, so ignore the 0, negative and undefined values. As the least ration is 43.33 which correspond to s2 we will replace it by x2 in the next table.

Step 5 The new solution is found by transforming the coefficients in the x2 column

Basic Variables

Key Column

Ratioz x1 x2 s1 s2 bi

1 -1 0 0 1 260

s1 0 -5 0 -3 1 -100

x2 0 4/6 1 0 1/6 260/6

Since all row(0) coefficients are not non-negative, i.e the solution is not optimal, so we start again from step 3.

The most negative value is -1 in row(0), so the column below x1 is key column.

The least ratio is 20 which corresponds to s1. So s1 is replace by x1 in the next table.

By applying Gaussian elimination method, we want to convert

Basic Variables

Key Column

Ratioz x1 x2 S1 S2 bi

1 0 0 3/5 4/5 280

x1 0 1 0 3/5 –1/5 20

x2 0 0 30 -12 17 900

• For a maximization problem having a mix of (≤, ≥, and =) constraints the simplex method itself does not change.• The only change is in transforming constraints to the standard equation form with appropriate supplemental variables.• Recall that for each (≥) constraint, a surplus variable is subtracted and an artificial variable is added to the left side of the constraint. For each (=) constraint, an artificial variable is added to the left side.

Maximization Problems with Mixed Constraints

• An additional column is added to the simplex tableau for each supplemental variable.

• Also, surplus and artificial variables must be assigned appropriate objective function coefficients (c, values)

• Surplus variables usually re-assigned an objective function coefficient of 0.

Maximization Problems with Mixed Constraints cont’d

• Artificial variables are assigned objective function coefficient of –M, where |M| is a very large number. This is to make the artificial variables unattractive in the problem.

• In any linear programming problem, the initial set of basic variables will consist of all the slack variables and all the artificial variables which appear in the problem.

Maximization Problems with Mixed Constraints cont’d

Maximize z = 8x1 + 6x2

subject to 2x1 + x2 ≥ 10

3x1 + 8x2 ≤ 96

x1, x2 ≥ 0

We have a maximization problem which has a mix of a (≤) constraint and a (≥) constraint.

Example

Rewriting the problem with constraints expressed as equations,

Maximize z = 8x1 + 6x2 + 0E1 – MA1 + 0S2

subject to 2x1 + x2 – E1 + A1 = 10

3x1 + 8x2 +S2 = 96

x1 , x2 E1, A1, S2 ≥ 0

Example cont’d

The initial tableau for this problem appears as in the following table:

Basic Variables

z x1 x2 E1 A1 S2 bi Row Number

1 –8 –6 0 M 0 0 (0)

A1 0 2 1 –1 1 0 10 (1)

S2 0 3 8 0 0 1 96 (2)

Example cont’d

• Note that the artificial variable is one of the basic variables in the initial solution.

• For any problem containing artificial variables, the row (0) coefficients for the artificial variable will not equal zero in the initial tableau.

• We need to make M equal to zero by

we get

Example cont’d

Basic Variables

z

Key Column

Transformed to zero Rowbi Number bi/aikx1 x2 E1 A1 S2

1 –8–2M –6–M M 0 0 –10M (0)

A1 0 2 1 –1 1 0 10 (1) 10/2 = 5*

S2 0 3 8 0 0 1 96 (2) 96/3 = 32

We want to transform

Example cont’d

Basic Variables

z

Key Column

Transformed to zero Rowbi Number Ratiox1 x2 E1 A1 S2

1 0 –4 –8 8+2M 0 80 (0)

x1 0 1 ½ –½ ½ 0 5 (1)

S2 0 0 13 3 -3 2 162 (2)

Example cont’d

By Gaussian elimination method, we want to transform

Basic Variables

z x1 x2 E1 A1 S2 bi Row Number

1 0 92 0 6M 16 1536 (0)

x1 0 6 16 0 0 2 192 (1)

E2 0 0 13 3 –3 2 162 (2)

Example cont’d

Example cont’d

In short form we can write.

The objective function is maximized at

Review

• The Simplex Method for Maximization problem with all ≤ constraints

• Maximization problem with mixed constraints

Next Lecture

• Minimization problems• Special phenomena

1. Alternative Optimal Solutions

2. No Feasible Solution

3. Unbounded Solutions• Examples